POJ 2386 Lake Counting (DFS)

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 32140 Accepted: 16061

Description

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors. 

Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M 

* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.

Sample Output

3

Hint

OUTPUT DETAILS: 

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

Source

USACO 2004 November

思路：

题意不多说。。不明白为什么在输入nm的时候用~scanf 就不行，按理说不应该啊，加上~scanf 也不执行函数体 就一直处于待输入状态，去掉就出结果了。想不明白。

另外一个就是如果出现  [Error] ld returned 1 exit status。就说明系统exe文件还在运行，建议强制关闭之后重启dev。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[105][105];int sum;int n,m;void dfs(int i,int j){int dir[8][2]={{0,1},{0,-1},{-1,0},{1,0},{1,-1},{-1,1},{1,1},{-1,-1}};a[i][j]=0;for(int w=0;w<8;w++){int dx=i+dir[w][0];int dy=j+dir[w][1];if(dx>=0&&dx<n&&dy>=0&&dy<m&&a[dx][dy]==1)dfs(dx,dy);}return;}int main(){cin>>n>>m; sum=0;for(int i=0;i<n;i++)for(int j=0;j<m;j++){char x;cin>>x;if(x=='.')a[i][j]=0;if(x=='W')a[i][j]=1;}for(int i=0;i<n;i++)for(int j=0;j<m;j++){if(a[i][j]==1){dfs(i,j);sum++;}}//for(int i=0;i<n;i++)//for(int j=0;j<m;j++)//{//cout<<a[i][j]<<" ";//if(j==m-1)cout<<endl;// } printf("%d\n",sum);return 0;}